
#109
Dec1103, 08:01 PM

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#110
Dec1103, 08:37 PM

P: 657

but note that you have dual vectors independently of the metric, and that the metric does not allow you to take inner products with a vector and a dual vector, but rather only two vectors (or also two dual vectors). so it makes no sense to talk about a vector and a dual vector being perpendicular, and that is what i was objecting to before. 



#111
Dec1103, 09:15 PM

P: 837

But what does whether they look straight or curved have to do with your ability to visualize them? You don't need a coordinate system in order to define a tangent space, and you don't need a tangent space in order to define a coordinate system. 



#112
Dec1103, 09:19 PM

P: 837





#113
Dec1103, 09:38 PM

P: 657

my question is this: if i have this orthonormal basis at each point, can i not integrate them to find the flow? can i not parametrize along these flows? will this parametrization not give me back the Kronecker delta as the components of the metric in these coordinates? that would be a contradiction if i were able to do that. so clearly it must not be possible. but on the other hand, i am under the impression that one can always integrate a vector field to find its flow. so this is my question. i think it has something to do with what you said, but i guess if you don t think so, well.... i must be awfully confused about a lot of things. 



#114
Dec1103, 09:52 PM

P: 837





#115
Dec1203, 08:34 AM

P: 657

should i infer that these integral curves don t exist? can you explain to me how come this integral curve does not have the vector it is an integral curve of as tangent? 



#116
Dec1203, 10:38 AM

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From QM, I remember that the inner product was a bra multiplying a ket (at least, that's how it was done in Shankar). I got the impression in QM that the bra and ket are literally different objects, even if it is the bra version and ket version of the same state. There was never a metric idea introduced, but orthogonality was still an important issue. From classical mechanics, I remember that any vector has two different kinds of components and two different kinds of bases, contravariant and covariant. The contravariant basis is like arrows, and the covariant basis is like a set of planes (stack of pancakes). Taking a dot product (which I'm still trying to decide if this is really the same as an "inner product") of two vectors is summing the products of the contravariant components of one of them with the covariant components of the other. If the dot product is zero, I interpretted it as the arrow being parallel to the planes in the set of planes. The dot product is maximum when the arrow is perpendicular to the set of planes (so that it passes through the maximum number of planes). Are you saying that my interpretations from classical mechanics are inappropriate? Should I think about it more in terms of the way I think about the inner product in QM? 



#117
Dec1203, 10:55 AM

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P: 2,327

But now I am confused about why one needs two coordinates instead of one. Is there no way to use just one coordinate to indicate points. I am having trouble shifting my mind from elements of a set needing only one index and points in a manifold needing more than one coordinate. (I am at this point going to assume that my analogy was correct that points:manifold::elements:set) In a vector space, I've got the idea of linear independence dictating how many independent vectors I will need in my basis to span the vector space, so the dimension is clear. I don't see the same indicator in a manifold. Can you explain the collection of functions a bit? 



#118
Dec1203, 11:08 AM

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#119
Dec1303, 04:29 AM

P: 657

so one way to think about an inner product is this: take two vectors in the vector space, feed them to the inner product machine, and get a number out. this is the better way to think of it, in my opinion. another way to think of it, is, take two vectors, then use the isomorphism determined by the inner product to find the dual of one of the vectors, and then feed the other vector to that dual, getting a number out. this latter is what you do in both descriptions you gave above. so in some sense, if you have an inner product, you can just not bother to distinguish between vectors and their dual. this is how you are thinking about vectors in classical: the only difference between a vector and a covector is whether it has raised or lowered indices. in QM, the only difference between a vector and a covector is whether it is a bra or a ket. so, im not going to tell you that this way of thinking about vector spaces is wrong, but i have learned that it is very useful to, at the very least, remain aware of when you are using your inner product, and for what purposes. you never know when you may need to change your inner product, and then this isomorphism will change, and all the dual vectors you had previously written down will be something different. better still, don t use the isomorphism at all, and only write down a vector if you actually mean a vector, and a covector if you mean a covector. 



#120
Dec1303, 05:00 AM

P: 657

i guess in a set theoretic sense, you don t actually need two indices. the cartesian product of two sets has the same cardinality, so you could count the elements of the new set with the same index, if you had a mind to, but this would muck up the other nice properties of the set, so it is much preferred to just stick to the double index concept. in other words, if you look at a manifold up closely enough, it looks like R^{n}. these local homeomorphisms are the charts of the manifold, they are the coordinates. there is no discontinuity. perhaps it would be useful to know the following fact: a smooth manifold, together with the tangent space at each point, considered as one larger set, is also a smooth manifold. this manifold is called the tangent bundle. 



#121
Dec1303, 05:20 AM

P: 657

this should be apparent to you. if you know that electric field is a vector field, and you know that electric field can be written [itex]\mathbf{E}=E_x\hat{\imath}+E_y\hat{\jmath}+E_z\hat{k}[/itex], so those i, j, and k are also vector fields. you can choose to work in polar coordinates, and thats all we are doing here. in short, there are various things you can do to define tangent vectors in an intrinsic way. a nice starting way is to define a tangent vector as an equivalence class of curves all of which have the same derivative at a point. with a little work, you see that this is canonically isomorphic to the space of first order differential operators on functions on the manifold. therefore i can simply say they are the same thing, and call a tangent vector on a manifold a differential operator. this is tremendously useful, though unfamiliar. 


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